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If I take a 12 V battery and connected its (+) terminal with the (+) wire of a load and connected the negative wire from the load to the ground (earthing). I have heard that earthing absorbs current in AC system (Used as return path). Do it is same on DC.

Is this circuit be close or open? Please give reason for your statement.

Would the LED Glow or not?

Thanks in Advance, Your response will be highly appreciated.

enter image description here

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    \$\begingroup\$ Ok, there's a lot of problems with this question. 1) You haven't connected the battery (-) to anything. 2) A 15 Watt-hour LED doesn't make sense. 3) Where do you measure a 1.5 V difference? 4) The way you phrase it makes it look like a homework question. 5) No one is going to tell you "more about A/C & D/C systems" unless you tell exactly what you want to know. \$\endgroup\$ – pipe Apr 19 '16 at 3:18
  • \$\begingroup\$ What is the other terminal of the battery connected to? (The easy way to answer this question would be to include a schematic of your circuit: Edit your question and hit Ctrl-M to start the schematic editor) \$\endgroup\$ – The Photon Apr 19 '16 at 3:18
  • \$\begingroup\$ Other terminal is left untouched. \$\endgroup\$ – Saad Apr 19 '16 at 3:29
  • \$\begingroup\$ You have edited the question, and it is still entirely unclear. \$\endgroup\$ – uint128_t Apr 19 '16 at 3:30
  • \$\begingroup\$ I calculated the Potential difference with a Digital Multimeter \$\endgroup\$ – Saad Apr 19 '16 at 3:31
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If your latest photo is the COMPLETE circuit, then you have NO path for any current to flow through the load. If the - side of the battery is connected to nothing (as shown in your photo) then no current can flow.

It makes NO difference whether any part of the circuit is connected to ground.

Would the circuit be complete or not?

No. The circuit is NOT complete.

Please give reason for your statement.

You need a complete path (loop) for the current to connect from BOTH sides of the source (the battery) to the load. You have only one side connected.

I have heard that earthing pushes out current in AC system. Do it is same on DC.

That makes no sense. For AC or for DC.

Would the LED Glow or not?

No

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  • \$\begingroup\$ Actually a battery won't do WORK unless or until it's both terminals are connected with each other having a load in between. \$\endgroup\$ – Saad Apr 19 '16 at 13:18
  • \$\begingroup\$ Batteries only work when we connect both terminals so that +ve Voltage & -ve Voltage get meet at a Load where they get converted into any other energy. (i.e Light energy from LED Bulb \$\endgroup\$ – Saad Apr 19 '16 at 13:20
  • \$\begingroup\$ battery is 0. But if you test it you actually make a path and current starts flowing from positive electrode having P.D: +12 V through your Meter towards the negative electrode having P.D -12 V. But rather batteries are printed to be positive 12 V as negative voltage in battery doesn't count. Am I right? If not please tell me the right. \$\endgroup\$ – Saad Apr 19 '16 at 15:02
  • \$\begingroup\$ A circuit must have a COMPLETE path from the battery + to the load, and then from the other side of the load back to the - side of the battery. If you break any part of this loop path, no current will flow and you have an incomplete circuit. When you use your meter, the meter itself provides the current path. When you remove the meter, the circuit is broken and no current will flow. \$\endgroup\$ – Richard Crowley Apr 19 '16 at 16:10
  • \$\begingroup\$ But if I place any other type of DC source in place of a battery, would the circuit be complete? \$\endgroup\$ – Saad Apr 20 '16 at 13:33

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